Question

A ball is launched into the sky at 19.6 feet per second from a 58.8 meter tall building. The equation for the ball’s height, h, at time t seconds is h = -4.9t2 + 19.6t + 58.8 . When will the ball strike the ground?

Answer 1

Answer:

Therefore,

The ball strike the ground at time , t = 6 s

Step-by-step explanation:

Given:

The equation for the ball’s height, h, at time t seconds is

$h=-4.9t^{2}+19.6t+58.8$

To Find:

time, t = ?   (When the ball strike the ground)

Solution:

When the ball strike the ground height, h will be ZERO,

So put h = 0 in above given equation for time,

$0=-4.9t^{2}+19.6t+58.8$

Which is a Quadratic Equation

Dividing throughout by - 4.9 we get

$0=t^{2}-4t-12$

On Factorizing, splitting the middle term we get

$t^{2}-6t+4t-12=0$

$t(t-6)+2(t-6)=0\\\\(t+2)(t-6)=0\\\\t=-2\ or\ t=6$

As time cannot be negative,

t = 6 s

Therefore,

The ball strike the ground at time , t = 6 s